Logarithms may look difficult to use, but just like exponents or polynomials, you just need to learn the correct techniques. You only need to know a couple basic properties to divide two logarithms of the same base, or to expand a logarithm that contains a quotient.

Steps

Method1

Dividing Logarithms by Hand

1

Check for negative numbers and ones. This method covers problems in the form logb⁡(x)logb⁡(a){\displaystyle {\frac {\log _{b}(x)}{\log _{b}(a)}}}. However, it does not work for a few special cases:[1]

The log of a negative number is undefined for all bases (such as log⁡(−3){\displaystyle \log(-3)} or log4⁡(−5){\displaystyle \log _{4}(-5)}). Write "no solution."

The log of zero is also undefined for all bases. If you see a term such as ln⁡(0){\displaystyle \ln(0)}, write "no solution."

The log of one in any base (log⁡(1){\displaystyle \log(1)}) always equals zero, since x0=1{\displaystyle x^{0}=1} for all values of x. Replace that logarithm with 1 instead of using the method below.

If the two logarithms have different bases, such as log3(x)log4(a){\displaystyle {\frac {log_{3}(x)}{log_{4}(a)}}}, and you cannot simplify either one into an integer, the problem is not feasible to solve by hand.

2

Convert the expression into one logarithm. Assuming you did not find any of the exceptions above, you can now simplify the problem into one logarithm. To do this, use the formula logb⁡(x)logb⁡(a)=loga⁡(x){\displaystyle {\frac {\log _{b}(x)}{\log _{b}(a)}}=\log _{a}(x)}.

This formula is the "change of base" formula, derived from basic logarithmic properties.

3

Calculate by hand if possible. Remember, to solve loga⁡(x){\displaystyle \log _{a}(x)}, think "a?=x{\displaystyle a^{?}=x}" or "What exponent can I raise a by to get x?" It's not always feasible to solve this without a calculator, but if you're lucky, you'll end up with an easily simplified logarithm.

Example 1 (cont.): Rewrite log2⁡(16){\displaystyle \log _{2}(16)} as 2?=16{\displaystyle 2^{?}=16}. The value of "?" is the answer to the problem. You may need to find it by trial and error:22=2∗2=4{\displaystyle 2^{2}=2*2=4}23=4∗2=8{\displaystyle 2^{3}=4*2=8}24=8∗2=16{\displaystyle 2^{4}=8*2=16}16 is what you were looking for, so log2⁡(16){\displaystyle \log _{2}(16)} = 4.

4

Leave the answer in logarithm form if you cannot simplify it. Some logarithms are very difficult to solve by hand. You'll need a calculator if you need the answer for a practical purpose. If you're solving problems in math class, your teacher most likely expects you to leave the answer as a logarithm. Here's another example using this method on a more difficult problem:

Convert this into one logarithm: log3⁡(58)log3⁡(7)=log7⁡(58){\displaystyle {\frac {\log _{3}(58)}{\log _{3}(7)}}=\log _{7}(58)}. (Notice that the 3 in each initial log disappears; this is true for any base.)

Rewrite as 7?=58{\displaystyle 7^{?}=58} and test possible values of ?:72=7∗7=49{\displaystyle 7^{2}=7*7=49}73=49∗7=343{\displaystyle 7^{3}=49*7=343}Since 58 falls between these two numbers, log7⁡(58){\displaystyle \log _{7}(58)} has no integer answer.

Leave your answer as log7⁡(58){\displaystyle \log _{7}(58)}.

Method2

Working with the Log of a Quotient

1

Start with a division problem inside a logarithm. This section helps you solve problems that include expressions in the form loga⁡(xy){\displaystyle \log _{a}({\frac {x}{y}})}.

For example, start with this problem:"Solve for n if log3⁡(276n)=−6−log3⁡(6){\displaystyle \log _{3}({\frac {27}{6n}})=-6-\log _{3}(6)}."

2

Check for negative numbers. The logarithm of a negative number is undefined. If x or y are a negative numbers, confirm that the problem has a solution before you continue:

If either x or y is negative, there is no solution to the problem.

If both x and y are negative, remove the negative signs using the property −x−y=xy{\displaystyle {\frac {-x}{-y}}={\frac {x}{y}}}

There are no logarithms of negative numbers in the example problem, so you can continue to the next step.

3

Expand the quotient into two logarithms. One useful property of logarithms is described by the formula loga⁡(xy)=loga⁡(x)−loga⁡(y){\displaystyle \log _{a}({\frac {x}{y}})=\log _{a}(x)-\log _{a}(y)}. In other words, the log of a quotient is always equal to the log of the numerator minus the log of the denominator.[2]

Use this to expand the left side of the example problem:log3⁡(276n)=log3⁡(27)−log3⁡(6n){\displaystyle \log _{3}({\frac {27}{6n}})=\log _{3}(27)-\log _{3}(6n)}

Substitute this back into the original equation:log3⁡(276n)=−6−log3⁡(6){\displaystyle \log _{3}({\frac {27}{6n}})=-6-\log _{3}(6)}→log3⁡(27)−log3⁡(6n)=−6−log3⁡(6){\displaystyle \log _{3}(27)-\log _{3}(6n)=-6-\log _{3}(6)}

4

Simplify the logarithms if possible. If any of the new logarithms in the expression have an integer answer, simplify them now.

The example problem has a new term: log3⁡(27){\displaystyle \log _{3}(27)}. Since 33 = 27, simplify log3⁡(27){\displaystyle \log _{3}(27)} to 3.

The full equation is now:3−log3⁡(6n)=−6−log3⁡(6){\displaystyle 3-\log _{3}(6n)=-6-\log _{3}(6)}

5

Isolate the variable. Just like any algebra problem, it helps to isolate the term with the variable on one side of the equation. Combine like terms whenever possible to simplify the equation.

In the example problem, the n is still trapped inside the term log3⁡(6n){\displaystyle \log _{3}(6n)}.In order to isolate the n, use the product property of logarithms: loga⁡(bc)=loga⁡(b)+loga(c){\displaystyle \log _{a}(bc)=\log _{a}(b)+\log {a}(c)}log3⁡(6n)=log3⁡(6)+log3⁡(n){\displaystyle \log _{3}(6n)=\log _{3}(6)+\log _{3}(n)}

Substitute this back into the full equation:log3⁡(6n)=9+log3⁡(6){\displaystyle \log _{3}(6n)=9+\log _{3}(6)}log3⁡(6)+log3⁡(n)=9+log3⁡(6){\displaystyle \log _{3}(6)+\log _{3}(n)=9+\log _{3}(6)}

7

Continue simplifying until you find the solution. Repeat the same algebra and logarithmic techniques to solve the problem. If there is no integer solution, use a calculator and round to the nearest significant figure.

References

To divide logarithms by hand, start by checking for negative numbers and ones. If you don’t find any exceptions to the standard rules, you can simplify the problem into 1 logarithm. Whenever possible, calculate the problems by hand, but, if need be, you can use a calculator to help. If you can’t simplify the problem, leave the answer in logarithmic form.